Find the sums given below:
(i) \(1046\dfrac{1}{2}\)
(ii) \(286\)
(iii) \(-8930\)
Recall: For an AP with first term \(a\), common difference \(d\), and \(n\) terms,
\[a_n = a + (n - 1)d, \quad S_n = \frac{n}{2}(a + a_n)\]
First, write the terms in decimal or fraction form:
\[7,\ 10\tfrac{1}{2} = 10.5,\ 14,\ldots,\ 84\]
First term: \(a = 7\).
Second term: \(10.5\). So the common difference is
\[d = 10.5 - 7 = 3.5\]
We know the last term \(a_n = 84\). Use \(a_n = a + (n - 1)d\):
\[84 = 7 + (n - 1)\cdot 3.5\]
\[(n - 1)\cdot 3.5 = 77\]
\[n - 1 = \frac{77}{3.5} = 22 \Rightarrow n = 23\]
Now use the sum formula:
\[S_{23} = \frac{23}{2}(7 + 84) = \frac{23}{2} \times 91\]
\[S_{23} = \frac{23 \times 91}{2} = \frac{2093}{2} = 1046\tfrac{1}{2}\]
So, the sum is \(1046\tfrac{1}{2}\).
Sequence: \(34, 32, 30, \ldots, 10\).
First term: \(a = 34\).
Common difference:
\[d = 32 - 34 = -2\]
Last term: \(a_n = 10\).
Use \(a_n = a + (n - 1)d\):
\[10 = 34 + (n - 1)(-2)\]
\[10 = 34 - 2(n - 1)\]
\[10 = 34 - 2n + 2 = 36 - 2n\]
\[2n = 36 - 10 = 26 \Rightarrow n = 13\]
Now find the sum:
\[S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2} \times 44\]
\[S_{13} = 13 \times 22 = 286\]
So, the sum is \(286\).
Sequence: \(-5, -8, -11, \ldots, -230\).
First term: \(a = -5\).
Common difference:
\[d = -8 - (-5) = -3\]
Last term: \(a_n = -230\).
Use \(a_n = a + (n - 1)d\):
\[-230 = -5 + (n - 1)(-3)\]
\[-230 = -5 - 3(n - 1)\]
\[-230 = -5 - 3n + 3 = -2 - 3n\]
\[-3n = -230 + 2 = -228 \Rightarrow n = 76\]
Now find the sum:
\[S_{76} = \frac{76}{2}(-5 + (-230)) = 38 \times (-235)\]
\[S_{76} = -8930\]
So, the sum is \(-8930\).